Mathematics Advanced • Year 11 • Module 4 • Lesson 3

The Number e and Natural Exponentials

Practise HSC-style writing on e^x, continuous compounding, and exponential modelling.

Master · Past-Paper Style

1. Short-answer questions

1.1 Evaluate e^1.5 and e^−0.5 to 3 decimal places. State the exact relationship between the two answers in one line. 3 marks Band 3

1.2 $20 000 is invested at 6% per annum compounded continuously. How many years (to 1 dp) until the investment doubles? 3 marks Band 4

1.3 The temperature of a metal cooling in still air is T(t) = 25 + 75 e^{−0.04 t}, with t in minutes and T in °C.
(a) State the ambient (room) temperature and the initial temperature.
(b) Find the time, to the nearest minute, at which the metal first cools below 50°C. 4 marks Band 4

Stuck on 1.3(b)? Isolate the exponential, take ln, solve for t.

2. Extended response

2.1 A bank advertises a savings product earning 8% per annum.
(a) For an initial deposit of $10 000, compute the balance after 5 years under (i) annual compounding, (ii) quarterly compounding, (iii) continuous compounding. Round each to the nearest dollar.
(b) Using the limit definition of e, namely e = lim_{n → ∞} (1 + 1/n)ⁿ, show formally that the continuous-compounding formula A = Pe^rt arises as the limit of the periodic-compounding formula A = P (1 + r/n)^nt as n → ∞.
(c) For an 8% rate over 5 years, comment on the size of the gap between quarterly compounding and continuous compounding. Why does this gap shrink in dollar terms even as compounding becomes "more frequent"? 7 marks Band 5-6

Explicit marking criteria

Part (a) — 3 marks

• 1 mark — annual: 10 000 × 1.08⁵ ≈ $14 693.

• 1 mark — quarterly: 10 000 × (1 + 0.08/4)²⁰ = 10 000 × (1.02)²⁰ ≈ $14 859.

• 1 mark — continuous: 10 000 × e^0.4 ≈ $14 918.

Part (b) — 3 marks

• 1 mark — rewrites (1 + r/n)^nt = [(1 + r/n)^n/r]^rt, substituting m = n/r so n/r → ∞ as n → ∞.

• 1 mark — applies (1 + 1/m)^m → e to conclude (1 + r/n)^n/r → e.

• 1 mark — therefore (1 + r/n)^nt → e^rt, hence A = Pe^rt.

Part (c) — 1 mark

• 1 mark — quarterly to continuous: $14 918 − $14 859 = $59 (small). The gap shrinks because each additional compounding step adds an ever-smaller marginal "interest on interest"; the sequence (1 + r/n)ⁿ converges to e^r, so further frequency offers diminishing returns.

Your response:

Stuck on (b)? The trick is to introduce m = n/r so the inside becomes the canonical (1 + 1/m)^m.

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What I'll revisit before next class:

Answers — sample responses + marking notes

1.1 — Evaluate e^1.5, e^−0.5 (3 marks)

Sample response. e^1.5 ≈ 4.482. e^−0.5 ≈ 0.607. Relationship: e^1.5 × e^−0.5 = e¹ = e ≈ 2.718, so the two values multiply to e (equivalently, e^−0.5 = 1/e^0.5 ≈ 1/1.649 ≈ 0.607).

Marking notes. 1 mark per correct value (using e^x button, not powers of 2.718). 1 mark for an explicit algebraic relationship — most commonly the product = e, or the ratio = e². Bare values without a relationship → 2/3.

1.2 — Doubling time at 6% continuous (3 marks)

Sample response. Doubling: 2P = Pe^{0.06 t} ⇒ 2 = e^{0.06 t} ⇒ 0.06 t = ln 2 ⇒ t = ln 2 / 0.06 ≈ 0.6931 / 0.06 ≈ 11.6 years.

Marking notes. 1 mark — sets up 2 = e^{0.06 t}. 1 mark — takes ln of both sides correctly. 1 mark — final value to 1 dp (≈ 11.6 years). Note the answer is independent of P — a clean Band-4 student writes "doubling time depends only on r".

1.3 — Cooling metal T(t) = 25 + 75 e^{−0.04 t} (4 marks)

Sample response.
(a) Ambient: T → 25°C as t → ∞ (asymptote). Initial: T(0) = 25 + 75 = 100°C.
(b) 50 = 25 + 75 e^{−0.04 t} ⇒ 25 = 75 e^{−0.04 t} ⇒ e^{−0.04 t} = 1/3 ⇒ −0.04 t = ln(1/3) = −ln 3 ⇒ t = ln 3 / 0.04 ≈ 1.0986 / 0.04 ≈ 27.5 min. The metal first cools below 50°C at t ≈ 28 minutes.

Marking notes. (a) 1 mark — ambient 25°C; 1 mark — initial 100°C. (b) 1 mark — isolates e^{−0.04 t} = 1/3 correctly; 1 mark — t ≈ 27.5 rounded to 28. Round-to-nearest students who write "t = 27" earn 0.5 (the metal is still at 50°C, not "first below").

2.1 — Extended response (7 marks): sample Band-6 response with annotations

Sample Band-6 response.

Part (a) — Compounding at 8% for 5 years.

(i) Annual: A = 10 000 × (1.08)⁵ = 10 000 × 1.46933 ≈ $14 693. [1 mark.]
(ii) Quarterly: A = 10 000 × (1 + 0.08/4)^{4 × 5} = 10 000 × (1.02)²⁰ ≈ 10 000 × 1.48595 ≈ $14 859. [1 mark.]
(iii) Continuous: A = 10 000 × e^{0.08 × 5} = 10 000 × e^0.4 ≈ 10 000 × 1.49182 ≈ $14 918. [1 mark.]

Part (b) — Deriving A = Pe^rt as a limit.

The periodic-compounding formula is A = P (1 + r/n)^nt. Let m = n/r, so n = mr. Then:

A = P (1 + 1/m)^mrt = P [(1 + 1/m)^m]^rt. [1 mark — substitution m = n/r and rewrite.]

As n → ∞ (and so m = n/r → ∞), the limit definition of e gives (1 + 1/m)^m → e. [1 mark.]

Therefore A → P · e^rt = P e^rt. [1 mark — concludes formula.]

Part (c). The quarterly-to-continuous gap is $14 918 − $14 859 = $59, which is small compared to the $5 000 of total earnings. The gap shrinks because the sequence (1 + r/n)ⁿ converges to the finite value e^r, so each doubling of n produces ever-smaller marginal gains — there is no infinite jackpot in "compounding more often". [1 mark.]

Total: 7/7.

Band descriptors for marker.

Band 3: Correct annual; struggles to express quarterly inside (1 + r/n)^nt. Skips (b). ≈ 2-3 marks.

Band 4: All three (a) values correct. (b) writes the limit definition without the substitution step; (c) acknowledges the gap is small but does not link to the convergence of (1 + 1/m)^m. ≈ 4-5 marks.

Band 5: Full (a). (b) executes the substitution m = n/r and identifies the limit but misses one algebra step. (c) gives a numerical gap and a qualitative reason. ≈ 5-6 marks.

Band 6: Full proof in (b) with each algebraic step justified; (c) uses the language of convergence ("each doubling of n produces ever-smaller marginal gains") and gives the explicit dollar gap. 7/7.